The problem asks for the values of $x$ at which the graph of the function:

$f(x) = \frac{2x^3}{3} - 4x^2 + 8$

has a horizontal tangent. To solve this, we need to find when the derivative $f'(x)$ is equal to zero, since a horizontal tangent occurs where the slope of the function is zero.

Step 1: Differentiate the function

Let's first calculate the derivative of $f(x)$:

$f(x) = \frac{2x^3}{3} - 4x^2 + 8$

Using basic differentiation rules:

$f'(x) = \frac{d}{dx} \left( \frac{2x^3}{3} \right) - \frac{d}{dx} \left( 4x^2 \right)$

$f'(x) = 2x^2 - 8x$

Step 2: Set the derivative equal to zero

To find when the tangent is horizontal, we solve:

$2x^2 - 8x = 0$

Factor the equation:

$2x(x - 4) = 0$

This gives us two solutions:

$x = 0 \quad \text{or} \quad x = 4$

Step 3: Conclusion

The function has horizontal tangents at $x = 0$ and $x = 4$. Therefore, the correct answer is:

$\boxed{\text{Para } x = 0 \text{ y } x = 4}$

This corresponds to the second option in the given choices.

Would you like more details on this solution, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. What is the geometric interpretation of a horizontal tangent line?
2. How do we identify critical points of a function using its derivative?
3. What does the second derivative tell us about the concavity of a function at these points?
4. How does the behavior of the function change at $x = 0$ and $x = 4$?
5. Can a function have more than one horizontal tangent in the same interval?

Tip: Always remember that horizontal tangents occur where the derivative equals zero.